I have the operator $T \in \mathcal L(\ell^2(\mathbb C))$ defined as $$ Tx_n = \frac {x_n} n \,. $$ I am supposed to show that $0$ is in the spectrum of $T$, $\sigma(T)$, knowing that $0$ is not an eigenvalue of $T$ and that $\sigma(T)$ is closed.
The spectrum of a linear bounded operator $T$ (on this particular space) was defined as \begin{equation}\label{eq:spectrum} \sigma(T) = \left\{ \lambda \in \mathbb C \mid (\lambda I - T)^{-1} \notin \mathcal L(\ell^2(\mathbb C)) \right\} \,. \tag{1} \end{equation}
An incomplete attempt
We know that $\sigma(T)$ is closed, so it contains all of its accumulation points. To show $0 \in \sigma(T)$, one approach might then be to show that $0$ is one of the accumulation points of $\sigma(T)$.
If this was the case, then there should be a sequence $(\sigma_n)\subset\sigma(T)$, for which $$ \Vert 0 - \sigma_n \Vert_2 = \Vert \sigma_n \Vert_2 = \left( \sum_{n=1}^\infty |\sigma_n|^2 \right)^{1/2} < \varepsilon, $$ for some large enough $n$. As $\sigma_n \in \sigma(T)$ for all $n$, by \eqref{eq:spectrum} we know that the operator $\sigma_n I - T$ is not bounded nor linear.
But then what? Does $0$ not being an eigenvalue step into the equation here somehow?
Well, we easily see that if $e^n_k=\delta_{n,k}$ (the Kronecker delta), then $Te^n=\frac{1}{n} e^n$. Accordingly, $\frac{1}{n}\in \sigma(T)$ for all $n\in\mathbb{N}$. Since $\sigma(T)$ is closed, we get that $0=\lim_{n\to\infty} \frac{1}{n}\in\sigma(T)$.
Alternatively, you can simply say that $T-0\cdot I=T,$ which is injective with $(T^{-1}x)_n=n x_n$ for $x\in T(\ell^2)$. Since $\| T^{-1} e^n\|_{\ell^2}=n,$ we see, directly, that $T^{-1}$ is not bounded.
A third approach is to show that $T$ is not surjective. This also isn't too bad as, for instance, the sequence given by $x_n=\frac{1}{n}$ is clearly not in the image of $T$.